1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitation force of a disk with a hole

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Hello, I have derived the equation for the gravitational force for a disk to be
    2Ggm/a^2(1-x/sqrt(a^2 - x^2) when an object is added on top of the system. My question is would the force still be somewhat similar if the disk now had a small hollow of radius c and from the center to the end of the disk the distance is d?


    3. The attempt at a solution
    Im assuming it will be and that the only difference I would make is that my radius is now (d-c) or my new area would now be da = 2 * pi * (d^2 - c^2) but im not sure if that logic makes sense.
     
  2. jcsd
  3. May 4, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    A sketch or at least a description of the variables would help.
    The expression has mismatching brackets.

    A disk with an empty interior can be described as sum of a larger disk with positive mass and a smaller disk with negative mass.
     
  4. May 5, 2017 #3
    Sorry! Here is a sketch. Im assumning if we have a regular disk then the gravitional force can be modeled by the force equation in the diagram!
    Tt7n5MV.png

    Now if we have a disk with an empty interior of radius b like this figure
    MuqfXnu.png

    Would the only thing that change in my equation would now be the radius? That the new radius would be (a -b) or would it be my area such that A = pi*b^2 - pi * a^2 or A=pi * (b-a)^2
     
  5. May 5, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    From your second post, I gather that is:
    ##\frac{2Ggm}{a^2}\left(1-\frac x{\sqrt{a^2 - x^2}}\right)##
    But that cannot be right. When x=0 the force should be 0.
     
  6. May 5, 2017 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    @haruspex: The force is discontinuous at x=0 for an ideal disk, the formula is for x>0.
    No, but you can model the force as sum of two disks, one with a positive and one with a negative mass.

    The square root in your formula cannot be correct - it becomes undefined for x>a.
     
  7. May 9, 2017 #6
    Does this mean the force is just F= F_actual_mass_with_radius_b - F_negative_mass_with_radius_a?
     
  8. May 9, 2017 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Gravitation force of a disk with a hole
  1. Disk with hole (Replies: 10)

  2. Disk With Hole Inertia (Replies: 4)

Loading...